Algebra Examples

$34^{3} - 27 h^{3}$

Step 1

Rewrite $27 h^{3}$ as ${(3 h)}^{3}$ .

$34^{3} - {(3 h)}^{3}$

Step 2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 34$ and $b = 3 h$ .

$(34 - (3 h)) (34^{2} + 34 (3 h) + {(3 h)}^{2})$

Step 3

Simplify.

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Step 3.1

Multiply $3$ by $- 1$ .

$(34 - 3 h) (34^{2} + 34 (3 h) + {(3 h)}^{2})$

Step 3.2

Raise $34$ to the power of $2$ .

$(34 - 3 h) (1156 + 34 (3 h) + {(3 h)}^{2})$

Step 3.3

Multiply $3$ by $34$ .

$(34 - 3 h) (1156 + 102 h + {(3 h)}^{2})$

Step 3.4

Apply the product rule to $3 h$ .

$(34 - 3 h) (1156 + 102 h + 3^{2} h^{2})$

Step 3.5

Raise $3$ to the power of $2$ .

$(34 - 3 h) (1156 + 102 h + 9 h^{2})$